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jeka57 [31]
1 year ago
13

Use the periodic table to choose the element that matches each description.

Chemistry
1 answer:
nika2105 [10]1 year ago
8 0

Using the periodic table to choose the element that matches each description include the following below:

  • Halogen: iodine
  • Group IIA: magnesium
  • Nonreactive: argon
  • Alkali metal: potassium.

<h3>What is a Periodic table?</h3>

This contains elements which are arranged according to the order of their atomic number in a tabular form. There are 18 groups which are the vertical columns present while there are 8 periods which are the horizontal rows present in the periodic table.

Example of an alkali metal is potassium while the non reactive ones include argon, neon etc. Examples of halogens include chlorine, iodine etc. are the ones which have seven electrons in their outer electron shells thereby just requiring one electron to achieve to obtain a stable octet configuration.

These are therefore the elements which match the descriptions provided in this case and is the most appropriate choice.

Read more about Periodic table here brainly.com/question/15987580

#SPJ1

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Give the oxidation state of the metal species in each complex. ru(cn)(co)4 -
kogti [31]
The given complex ion is as follow,

                                              [Ru (CN) (CO)₄]⁻

Where;
            [ ]  =  Coordination Sphere

            Ru  =  Central Metal Atom  =  <span>Ruthenium

            CN  =  Cyanide Ligand

            CO  =  Carbonyl Ligand

The charge on Ru is calculated as follow,

                               Ru + (CN) + (CO)</span>₄  =  -1
Where;
            -1  =  overall charge on sphere

             0  =  Charge on neutral CO

            -1  =  Charge on CN

So, Putting values,


                               Ru + (-1) + (0)₄  =  -1

                               Ru - 1 + 0  =  -1

                               Ru - 1  =  -1

                               Ru  =  -1 + 1

                               Ru  =  0
Result:
          <span>Oxidation state of the metal species in each complex [Ru(CN)(CO)</span>₄]⁻ is zero.
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Which of the following represents the electron configuration of rhodium
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3 years ago
An amphoteric salt is one that contains an anion that can act as either an acid or a base in water. Baking soda, NaHCO3, is an e
LenKa [72]

Answer:

pH = 8.34

Explanation:

The equilbriums of the amphoteric HCO₃⁻ (Ion of NaHCO₃) are:

H₂CO₃ ⇄ <em>HCO₃⁻</em> + H⁺ Ka1 <em>-Here, HCO₃⁻ is acting as a base-</em>

<em>HCO₃⁻</em>⇄ CO₃²⁻ + H⁺ Ka2 <em>-Here, is acting as an acid-</em>

Where Ka1 = 4.3x10⁻⁷ and Ka2 = 4.8x10⁻¹¹. As pKa = -log Ka:

pKa1 = 6.37; pKa2 = 10.32

As the pH of amphoteric salts is:

pH = (pKa1 + pKa2) / 2

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3 years ago
The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. A 1.376 g sample of e
hoa [83]

Answer:

The answer to your question is: C₄H₁₀O

Explanation:

Data

          CxHyOz

mass sample : 1.376 g

mass CO₂ = 3.268 g

mass H₂O = 1.672 g

Process

Reaction

                      CxHyOz  + O₂ ⇒   CO₂  +  H₂O

1.- Calculate the moles and mass of carbon

Molecular mass CO₂ = 44g

                      44 g of CO₂ --------------  12 g of C

                      3.268 g of CO₂  --------    x

                         x = (3.268 x 12) / 44

                        x = 0.891 g of Carbon

                       12 g of carbon -----------  1 mol

                       0.891 g of C     ----------   x

                       x = (0.891 x 1) / 12

                       x = 0.0743 moles of carbon

2.- Calculate the moles and mass of hydrogen

                      18 g of water --------------- 2 g of H

                      1.672 g of H₂O ------------  x

                      x = (1.672 x 2) / 18

                      x = 0.186 g of hydrogen

                      1 g of hydrogen ------------  1 mol of H

                      0.186 g of H       ------------  x

                      x = (0.186 x 1) / 1

                      x = 0.186 moles of H

3.- Calculate the mass of Oxygen and its moles

Mass of Oxygen = 1.376 - 0.891 - 0.186

                           = 0.299 g of O₂

Moles of Oxygen

                             16 g of Oxygen ---------------- 1 mol

                             0.299 g of O    -----------------  x

                             x = (0.299 x 1) / 16

                             x = 0.019 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon         0.0743/ 0.019 = 3.9 ≈ 4.0

Hydrogen     0.186/ 0.019 = 9.7 = 10

Oxygen         0.019/ 0.019 = 1

5.- Write the empirical formula

                              C₄H₁₀O                  

4 0
3 years ago
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